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I have to use a binomial expansion to evaluate $1/\sqrt{4.2}$ to $5$ decimal places. The answer from a calculator is $0.48795$ but I get $0.48202$, so I'm doing something wrong. I've also checked my calculations on a spreadsheet so the problem is with my technique, not arithmetic. Here's what I'm doing:

Using

$$ (x + y)^{-n} = x^{-n}\sum_{k=0}^\infty {^{-n}}C_k \left( \frac y x \right)^k $$ I need $$ (4 + 0.2)^{-1/2} = 4 ^{-1/2} \sum_{k=0}^\infty {^{-n}}C_k \left( \frac {0.2} {4} \right)^k $$

$$ = \frac1 2 \sum_{k=0}^\infty {^{-n}}C_k \left( \frac {1} {2} \right)^k \left( \frac {1} {10} \right)^k $$ Using the recurrence relation $$ ^nC_{k+1} = \frac{n-k}{k+1} {^n}C_k $$ I caluculate

$^{-1/2}C_0=1$;

$^{-1/2}C_1=-3/4$;

$^{-1/2}C_2=5/8$;

$^{-1/2}C_3=-35/64$

So the evaluation should be: $$ \frac 1 2\left( 1 + \frac {-3}{4}.\frac1 2.\frac 1 {10} + \frac {5}{8}.\frac1 4.\frac 1 {100} + \frac {-35}{64}.\frac1 8.\frac 1 {1000} \right) $$

but this is incorrect, as described above. It is enough terms because the last one is $-0.0000342$.

What am I doing wrong?

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${}^{-1/2}C_1=\frac{-\frac12-0}{0+1}\cdot {}^{-1/2}C_0$ etc.

Also, the last term should be $<10^{-5}$ to be sure about the 5 places.

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  • $\begingroup$ Ah! I used k + 1 where I should have used k. Thanks. $\endgroup$ – Luigi Plinge Oct 7 '12 at 9:31
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Your calculation of $\binom{-1/2}1$ is off. For non-negative integer $k$

$$\binom{x}k=\frac{x^{\underline{k}}}{k!}=\frac{x(x-1)(x-2)\dots(x-k+1)}{k!}\;,$$

so

$$\begin{align*} \binom{-1/2}0&=1\\ \binom{-1/2}1&=\frac{-1/2}{1!}=-\frac12\;,\\ \binom{-1/2}2&=\frac{\left(-\frac12\right)\left(-\frac32\right)}{2!}=\frac38\;,\text{ and}\\ \binom{-1/2}3&=\frac{\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)}{3!}=-\frac5{16}\;. \end{align*}$$

In general

$$\begin{align*} \binom{-1/2}k&=(-1)^k\frac{1\cdot3\cdot5\cdot\ldots\cdot(2k-1)}{2^kk!}\\ &=(-1)^k\frac{(2k)!}{\left(2^kk!\right)^2}\\ &=(-1)^k\frac1{4^k}\binom{2k}k\;. \end{align*}$$

You also need to go a little further to be sure of five-place accuracy.

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  • $\begingroup$ BTW, what's that underline under the k called? $\endgroup$ – Luigi Plinge Oct 7 '12 at 9:43
  • $\begingroup$ @Luigi: That’s a falling power or falling factorial; other notations that I’ve seen are $x^{k\downarrow}$ and $(x)_k$. The rising version is $x^{\overline{k}}=x(x+1)(x+2)\dots(x+k-1)=x^{k\uparrow}=x^{(k)}$. $\endgroup$ – Brian M. Scott Oct 7 '12 at 9:45

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